Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode * constructSubtree(vector<int> &preOrder, int preStart, int preEnd, vector<int> &inOrder, int inStart, int inEnd){ // if(preStart < preEnd) return NULL ; int temp = preOrder[preStart] ; TreeNode *root = new TreeNode(temp) ; if(preStart == preEnd) return root ; int i; for(i = inStart ; i<= inEnd ;i++) if(inOrder[i] == temp ) break; root->left = preStart+1 <= preStart+i-inStart ? constructSubtree(preOrder, preStart+1, preStart+i-inStart, inOrder, inStart, i-1) : NULL; root->right = preStart+i-inStart+1 <= preEnd ? constructSubtree(preOrder, preStart+i-inStart+1, preEnd, inOrder, i+1, inEnd) : NULL; return root ; } TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { // Start typing your C/C++ solution below // DO NOT write int main() function if(preorder.size() == 0 ) return NULL; TreeNode *root = new TreeNode(preorder[0]) ; if(preorder.size() == 1 ) return root ; int i; int temp = preorder[0] ; for(i = 0; i< inorder.size(); i++) if(inorder[i] == temp) break ; root->left = 1 <= i ? constructSubtree(preorder, 1,i,inorder, 0, i-1) : NULL ; root->right = i+1 <= preorder.size()-1 ? constructSubtree(preorder, i+1,preorder.size()-1, inorder, i+1, inorder.size() -1): NULL ; return root; } };